Graphs of equations of the form ax + by = k

An equation of the form \(ax+by=k\), where \(a\), \(b\) and \(k\) are constants and \(a\) and \(b\) are not both zero, is a linear equation in two variables. Its graph is a straight line. When \(b\ne0\) it can be rearranged into the form \(y=mx+c\) that you met in Secondary 1.

Why it is a straight line: rearranging gives an equivalent equation, satisfied by exactly the same pairs \((x,y)\). For example,

\(\begin{aligned} 5x+4y&=20 \\ 4y&=20-5x \\ y&=-\dfrac{5}{4}x+5 \end{aligned}\)

so \(5x+4y=20\) is the straight line with gradient \(-\dfrac{5}{4}\) and \(y\)-intercept \(5\).

Method

1. Choose three values of \(x\) spread across the required range.
2. Substitute each value into \(ax+by=k\) and solve for \(y\), or rearrange to \(y=mx+c\) first. Record the pairs in a table.
3. Plot the points using the given scales and join them with a ruler. Two points fix the line; the third point checks your working.
4. A point lies on the line exactly when its coordinates make the left-hand side equal to the right-hand side. Use this to test a point or to find a missing coordinate.

Example

Draw the graph of \(5x+4y=20\) for \(-4\le x\le8\). The coefficient of \(y\) is \(4\), so choose values of \(x\) that are multiples of \(4\) and every \(y\) is a whole number.

When \(x=-4\): \(5(-4)+4y=20\), so \(4y=40\) and \(y=10\). When \(x=0\): \(4y=20\), so \(y=5\). When \(x=4\): \(20+4y=20\), so \(y=0\). When \(x=8\): \(40+4y=20\), so \(y=-5\).

Does \((-8,15)\) lie on the line? \(5(-8)+4(15)=-40+60=20\), which equals the right-hand side, so it does. Does \((2,3)\)? \(5(2)+4(3)=22\ne20\), so it does not.

Check: put a negative value of \(x\) in brackets, as in \(5(-4)=-20\). If your three points do not lie in a straight line, one of the calculated values is wrong; recalculate before drawing.